# Use the limit definition to find the slope of the tangent line at a point

Use the limit definition to find the slope of the tangent line to the graph of $f(x)=\sqrt{8x+1}$ at the point $(6,7)$.

I sort of get how to do this but the square root is making it difficult. I know there is way but I don't remember. Can someone show how to solve this problem?

• Welcome to MSE. Could you show us what you have tried so far? This will make it easyer for people to see where you get stuck. Also, as a new user you might want to read the tutorial. – gebruiker Jul 9 '14 at 19:38

$$\sqrt {u} - \sqrt {v} \;\; = \;\; \frac{\sqrt {u} - \sqrt {v}}{1} \cdot \frac{\sqrt {u} + \sqrt {v}}{\sqrt {u} + \sqrt {v}} \;\; = \;\; \frac{u-v}{\sqrt {u} + \sqrt {v}}$$
In your case, $u$ will be $8(6+h) + 1$ and $v$ will be $8\cdot6 + 1.$
\begin{align} f'(6)&=\lim\limits_{h\rightarrow0}\frac{f(6+h)-f(6)}{h} \\&=\lim\limits_{h\rightarrow0}\frac{\sqrt{8(6+h)+1}-7}{h} \\&=\lim\limits_{h\rightarrow0}\frac{\sqrt{8(6+h)+1}-7}{h}\times\frac{\sqrt{8(6+h)+1}+7}{\sqrt{8(6+h)+1}+7} \\&=\lim\limits_{h\rightarrow0}\frac{8(6+h)+1-49}{h(\sqrt{8(6+h)+1}+7)} \\&=\lim\limits_{h\rightarrow0}\frac{8h}{h(\sqrt{8(6+h)+1}+7)} \\&=\lim\limits_{h\rightarrow0}\frac{8}{\sqrt{8(6+h)+1}+7} \\&=\frac{8}{14}\\&=\frac{4}{7} \end{align}